# Mathematics Questions and Answers – Trigonometric Equations – 1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Trigonometric Equations – 1”.

1. The solutions of a trigonometric equation for which ___________ are called principal solutions.
a) 0 < x < 2π
b) 0 ≤ x < π
c) 0 ≤ x < 2π
d) 0 ≤ x < nπ

Explanation: The solutions of a trigonometric equation for which 0 ≤ x < 2π are called principal solutions. x should lie between 0 and 2π except 2π.

2. The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the principal solution.
a) True
b) False

Explanation: The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution. n is used in general solution only.

3. Find the principal solutions of the equation cos x = 1/2.
a) π/6, 5π/6
b) π/3, 5π/3
c) π/3, 2π/3
d) 2π/3, 5π/3

Explanation: Since cos x is positive in 1st and 4th quadrant, so there are two principal solutions to above equation i.e. x = π/3, 2π-π/3. So, x = π/3, 5π/3.
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4. Find the principal solutions of the equation tan x = – √3.
a) π/6, 5π/6
b) π/3, 5π/3
c) π/3, 2π/3
d) 2π/3, 5π/3

Explanation: Since tan x is negative in 2nd and 4th quadrant so, there are two principal solutions to above equation i.e. x = π – π/3, 2π – π/3. So, x=2π/3, 5π/3.

5. Find the principal solutions of the equation sec x = -2.
a) 2π/3, 4π/3
b) 2π/3, 5π/3
c) 4π/3, 5π/3
d) π/3, 5π/3

Explanation: Since sec x is negative in 2nd and 3rd quadrant so, there are two principal solutions to above equation i.e. x = π – π/3, π + π/3. So, x=2π/3, 4π/3.
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6. Find the principal solutions of the equation sin x = -1/√2.
a) π/4, 3π/4
b) 3π/4, 5π/4
c) 3π/4, 7π/4
d) 5π/4, 7π/4

Explanation: Since sin x is negative in 3rd and 4th quadrant so, there are two principal solutions to above equation i.e. x = 2π – π/4, 2π + π/4. So, x=5π/4, 7π/4.

7. Find the principal solutions of the equation cosec x=2.
a) π/6, 5π/6
b) π/6, 7π/6
c) 5π/6, 11π/6
d) 5π/6, 7π/6

Explanation: Since cosec x is positive in 1st and 2nd quadrant, so there are two principal solutions to above equation i.e. π/6, π – π/6. So, x=π/6, 5π/6.

8. Find the principal solutions of the equation cot x=1/√3.
a) π/3, 4π/3
b) 2π/3, 5π/3
c) 4π/3, 5π/3
d) π/3, 5π/3

Explanation: Since cot x is positive in 1st and 3rd quadrant, so there are two principal solutions to above equation i.e. π/3, π + π/3. So, x=π/3, 4π/3.

9. Find general solution to equation sin x = 1/2.
a) x = nπ + (-1)n π/3
b) x = nπ + (-1)n π/6
c) x = nπ + (-1)n 2π/3
d) x = nπ + (-1)n 5π/6

Explanation: sin x= 1/2
sin x = sin π/6
x = nπ + (-1)n π/6.

10. Find general solution to equation cos x = – 1/√2.
a) x = 2nπ±7π/4
b) x = 2nπ±3π/4
c) x = 2nπ±π/4
d) x = 2nπ±π/3

Explanation: cos x = – 1/√2
cos x = – cos π/4 = cos (π- π/4) = cos 3π/4
So, x = 2nπ±π/4.

11. Find general solution to equation cot x = √3.
a) x = nπ + π/3
b) x = nπ + π/6
c) x = nπ + 2π/3
d) x = nπ + 5π/6

Explanation: cot x = √3
tan x = 1/√3
tan x = tan π/6
x = nπ + π/6.

12. Solve: tan x = cot x
a) x = nπ/2 + (π/4)
b) x = nπ + (3π/4)
c) x = nπ + (π/4)
d) x = nπ/2 + (3π/4)

Explanation: tan x = cot x
tan x = cot (π/2 – x)
x = nπ + (π/2 – x)
2x = nπ + (π/2)
x = nπ/2 + (π/4).

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